To solve the problem, we need to find the rate constant (k) for the first-order reaction given the initial and final pressures. Here’s the step-by-step solution:
### Step 1: Understand the Reaction
The reaction is given as:
\[ A(g) \rightarrow 4B(g) + 3C(g) \]
### Step 2: Determine Initial and Final Pressures
- Initial total pressure, \( P_0 = 100 \, \text{mm Hg} \)
- Final total pressure after 20 minutes, \( P_t = 40 \, \text{mm Hg} \)
### Step 3: Calculate Change in Pressure
The change in pressure can be calculated as:
\[ \Delta P = P_0 - P_t = 100 \, \text{mm Hg} - 40 \, \text{mm Hg} = 60 \, \text{mm Hg} \]
### Step 4: Relate Pressure Change to Reaction Stoichiometry
From the stoichiometry of the reaction:
- For every 1 mole of A that reacts, it produces 4 moles of B and 3 moles of C, resulting in a total increase of 7 moles (4 + 3) for every mole of A consumed.
- Therefore, the decrease in pressure due to the consumption of A can be expressed as:
\[ \Delta P = P_0 - P_t = 7 \times \Delta P_A \]
Where \( \Delta P_A \) is the pressure change due to the consumption of A.
### Step 5: Calculate the Change in Pressure of A
From the above relationship:
\[ 60 \, \text{mm Hg} = 7 \times \Delta P_A \]
Thus,
\[ \Delta P_A = \frac{60 \, \text{mm Hg}}{7} \approx 8.57 \, \text{mm Hg} \]
### Step 6: Calculate the Remaining Pressure of A
The pressure of A at time \( t \) is:
\[ P_A = P_0 - \Delta P_A = 100 \, \text{mm Hg} - 8.57 \, \text{mm Hg} \approx 91.43 \, \text{mm Hg} \]
### Step 7: Use the First-Order Rate Equation
For a first-order reaction, the rate constant \( k \) can be calculated using the formula:
\[ k = \frac{2.303}{t} \log \left( \frac{P_0}{P_t} \right) \]
### Step 8: Convert Time to Seconds
Given time \( t = 20 \) minutes:
\[ t = 20 \times 60 = 1200 \, \text{seconds} \]
### Step 9: Substitute Values into the Rate Constant Equation
Now substituting the values into the equation:
\[ k = \frac{2.303}{1200} \log \left( \frac{100}{40} \right) \]
### Step 10: Calculate the Logarithm
Calculate the logarithm:
\[ \log \left( \frac{100}{40} \right) = \log(2.5) \approx 0.3979 \]
### Step 11: Final Calculation of k
Substituting the logarithm back into the equation:
\[ k = \frac{2.303}{1200} \times 0.3979 \]
\[ k \approx \frac{0.916}{1200} \approx 7.63 \times 10^{-4} \, \text{s}^{-1} \]
### Final Answer
The rate constant \( k \) is approximately:
\[ k \approx 7.63 \times 10^{-4} \, \text{s}^{-1} \]
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